Question

Suppose `0< a,b,c < 1` and `ab + bc + ca = 1`. Find the minimum value of `a + b + c + abc`?

Answer 1

`a+b+c+abc≥2`

Explanation:

By the arithmetic mean-geometric mean-harmonic mean inequality, `(a+b+c+abc)/4≥root(4)(a^2b^2c^2)≥4/(1/a+1/b+1/c+1/(abc))`, or `a+b+c+abc≥4sqrt(abc)≥16/(1/a+1/b+1/c+1/(abc))`.

Since `16/(1/a+1/b+1/c+1/(abc))=16/((bc+ac+ab+1)/(abc))` and `ab+bc+ac=1`, `a+b+c+abc≥4sqrt(abc)≥8abc`.

From `4sqrt(abc)≥8abc`, we can conclude that `abc≤1/4`. Then, `a+b+c+abc≥2`.

Answer 2

See below.

Explanation:

One way to elude the inequality restrictions is with a change of variable so making

`{(a->(sinalpha+1)/2),(b->(sinbeta+1)/2),(c->(singamma+1)/2):}`

the problem now in the variables `alpha, beta, gamma` has only equality restrictions and then we can approach it with the usual lagrange multipliers technique.

`f(alpha,beta,gamma) = 1/8(sinalpha+1)(sinbeta+1)(singamma+1)`
`g(alpha,beta,gamma)=1/4((sinalpha+1)(sinbeta+1)+(sinalpha+1)(singamma+1)+(sinbeta+1)(singamma+1))-1`

with the lagrangian

`L(alpha,beta,gamma,lambda)=f(alpha,beta,gamma)+lambda g(alpha,beta,gamma)`

The stationary point conditions are obtained by solving

`grad L=0`

or

#{(1/8 Cos alpha(5 + 4 lambda + Sin gamma + 2 lambda Sin gamma + Sin beta (1 + 2 lambda + Sin gamma))=0),(1/8 Cosbeta (5 + 4 lambda + Singamma + 2 lambda Singamma + Sinalpha(1 + 2 lambda + Singamma))=0),(1/8 Cosgamma(5 + 4 lambda + Sinbeta + 2 lambda Sinbeta + Sinalpha(1 + 2 lambda + Sinbeta))=0),(1/4 (2 Singamma + Sinbeta (2 + Singamma) + Sinalpha (2 + Sinbeta + Singamma)-1)=0):}#

Solving this system of equations with the help of Wolfram Alpha we get at

`((a,b,c,f(a,b,c)),(0, 1, 1,2),(1,0,1,2),(1,1,0,2),(1/sqrt[3], 1/sqrt[3], 1/sqrt[3],10/(3 sqrt[3])))`

Here `f(a,b,c) = a + b + c + a b c`

All this trouble could be avoided by observing the problem symmetry, so guessing `a=b=c`